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Re: Is this possible? Residue computation leads to complexInfinity?


Your Residue expression is unevaluated (Residue appears in output) since you 
left out part of the Residue syntax. However, your argument was evaluated to 
ComplexInfinity so z must already have had a value.

Clear[z];
FullSimplify[
  Residue[(I*z)/(1-(2+4*a)*z^2+z^4),
    {z,Sqrt[a+1]-Sqrt[a]}],a>0]

(I*(Sqrt[1 + 1/a] - 1))/(8*(-a + Sqrt[a^2 + a] - 1))


Bob Hanlon

> 
> From: "kiki" <lunaliu3 at yahoo.com>
To: mathgroup at smc.vnet.net
> Date: 2005/09/25 Sun AM 02:36:33 EDT
> Subject: [mg60725] [mg60719]  Is this possible? Residue computation leads to 
complexInfinity?
> 
> In[15]:=
> \!\(FullSimplify[Residue[\(\[ImaginaryI]\ z\)\/\(1 - \((2 + 4\ a)\)\ z\^2 + 
> z\
> \^4\), \@\(a + 1\) - \@a], \ a > 0]\)
> 
> Out[15]=
> \!\(Residue[ComplexInfinity, \(-\@a\) + \@\(1 + a\)]\) 
> 
> 
> 


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